An electric charge Q is distributed uniformly along a thíck and enormously long conducting wire with radius R and length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire) Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtain A- (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: Genc On the other hand, the Gaussian surface inside the wire is given by A- Using Equation 1, the electric field in simplified form is E-
An electric charge Q is distributed uniformly along a thíck and enormously long conducting wire with radius R and length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire) Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtain A- (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: Genc On the other hand, the Gaussian surface inside the wire is given by A- Using Equation 1, the electric field in simplified form is E-
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Transcribed Image Text:An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius R and
length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside
and outside the wire)
Solution
To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as
We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot
product to obtain
A =
(Equation 1)
Case 1: Inside the wire
Since, r falls inside the wire, then all the enclosed charge must be:
Genc =
On the other hand, the Gaussian surface inside the wire is given by
A =
Using Equation 1, the electric field in simplified form is
E =
Transcribed Image Text:Case 2: Outside the wire
Since, r falls outside the wire, then, all the charge must be enclosed, thus
qenc =
On the other hand, the Gaussian surface outside the wire is given by
A =
Using Equation 1, the electric field in simplified form is
E =
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